A Note on Computation of a Ray Bending Path

Transcription

1 A Note on Computation of a Ray Bending Path Jerry Tessendorf August 7, starting point Index of refraction spatial variations within a volume induce bending of the ray marching path that light takes through the participating medium. The driving equations for this bending describes the shape of the path xs and direction of propation ˆns as functions of the arclength s along the path. Both are modified by variations in the index of refraction n via the equations d xs dˆns = ˆns 1 = s ˆns ˆns s where s is the gradient of the log of the index of refraction: s = log n xs 3 The direction of propagation is also the tangent of the path. This sets up an initial value problem: Given the values at the initial arclength of x = x 0 and ˆn = ˆn 0, compute from these equations the position and direction for arclengths s > of interest. Here we consider a situation in which the the gradient vector is constant along a stretch of arclength of interest. There are two situations in which this occurs, and many more may be relevant. One of them is when the volume is modeled as a stack of layers and the gradient is constant within each layer. Another is the volume rendering ray march process with varying with each ray march step, but constant along the length of a single step. This latter situation is consistent with other assumptions in ray marching, which assumes that the material density and color are also constant over the distance of a single ray march step. In the next section, equation in solved exactly under the constant gradient assumption by expressing ˆns as a rotation about an axis. The path is found in section 3 by integrating equation 1 exactly. 1

2 ray bending as a direction rotation Ray bending alters the direction of the path tangent, while preserving the fact that the tangent vector is a unit vector. This means that the alteration of the path tangent can be represented by a rotation of the tangent vector. A general rotation is described by a unit vector ˆαs for the axis of rotation, and an angle αs for the amount of rotation about the axis. Using these two quantities, the rotation is expressible as ˆns = ˆn 0 cos αs + ˆαs ˆn 0 ˆαs 1 cos αs + ˆn 0 ˆαs sin αs 4 along with the initial condition α = 0. Ray bending with a constant provides a natural rotation axis: the path is contained in a plane defined by the two vectors ˆn 0 and ˆ = /, so the axis of rotation is constat and perpendicular to both of these: ˆα = ˆn 0 ˆ ˆn 0 ˆ 5 Because of this choice of rotation axis, the second term on the RHS of equation 4 is zero, and the third term has the vector ˆn 0 ˆα = ˆn 0 ˆ ˆn 0 ˆ ˆn 0 ˆ 6 which is a unit vector: ˆn 0 ˆα = 1 ˆn 0 ˆ ˆn 0 ˆ = 1 7 This brings the expression for the path tangent to ˆns = ˆn 0 cos αs + ˆn 0 ˆα sin αs 8 and the problem of describing the ray bending path tangent has been reduced to solving for the angle αs. To solve for αs, this expression for ˆns can be inserted into equation. The derivative on the LHS becomes dˆns = dαs { ˆn 0 sin αs + ˆn 0 ˆα cos αs} 9 Since equation has reduced to a differential equation for a single degree of freedom, taking the inner product of both sides with ˆn 0 simplifies the expressions. The LHS becomes ˆn 0 dˆns = dαs sin αs 10

3 and the RHS side is { ˆn 0 s ˆns ˆns s } = sin αs producing the differential equation dαs = ˆn 0 sin αs + { } ˆn 0 sin αs ˆn 0 ˆα cos αs 11 ˆn 0 ˆα cos αs 1 This expression can be solved for α by setting up the integral equation α 0 dx ˆn 0 ˆα cos x ˆn 0 = s 13 sin x This integral has a closed form expression 1 x + β αs log tan = s 0 ˆn 0 ˆ + ˆn 0 ˆα 14 where ˆn 0 ˆ = cos β. Using ˆn 0 ˆ + ˆn 0 ˆα =, 15 the solution is and β αs tan = tanβ/ e s s0 16 Using some identities, this can be manipulated into the form tanαs/ = tanβ/ 1 e s s0 1 + tan β/ e s s0 17 urther identity manipulation gives 1 + tan β/ e s s0 cosαs/ = cosβ/ tan β/ e s s0 e s s0 1 sinαs/ = sinβ/ tan β/ e s s0 cos αs = cos β + 4 sin β/ e s s0 + sin β/ tan β/ 1 s s0 e 1 + tan 1 β/ e s s0 sin αs = sin β 1 e s s0 1 + tan β/ e s s0 1 + tan β/ e s s0 0 This is a complete solution for the tangent of the ray bending path in a region of constant gradient. 1 Grahteyn and Ryzhik.557.4, pg

4 3 integrating the path The ray bending path xs follows from integrating equation 1 using equations 8, 1, and : xs = x 0 + ˆn 0 cos αs + ˆn 0 ˆα sin αs 3 These integrals have the generic form dt A + B e t + C e t 1 + D e t 4 for appropriate values of A, B, C, D. Using some integral identities this evaluates to dt A + B e t + C e t 1 + D e t = A t log 1 + D e t B + D tan 1 D e t + C D log 1 + D e t 5 Assembling all of the algebra that follows from these expressions, cos αs = cos β 1 + tan β/ s + log 1 + tan β/ e s s0 and + sin β sin αs = sin β sin β tan 1 tanβ/ e s s0 β/ tan β/ s + log 1 + tan β/ e s s0 1 tan β/ tanβ/ tan 1 tanβ/ e s s0 β/ 7 igure 1 shows these two functions, plotted parametrically by varying the value of s, for a range of values of β. 4 curved path intersection with a plane We can evaluate the problem of computing the point of intersection of a curved ray with a plane. The plane is characterized by the implicit equation Grahteyn and Ryzhik.313.1, pg. 9;.14.1, pg. 60 x x P ˆn P = 0 8 4

5 10 β = 0 π/ 8 6 s 0 sin αs s cos αs s0 igure 1: The two components of the ray bending path, plotted parametrically. Each line is a different value of β, which ranges in value from 0 to π/. 5

6 where x P is a reference point on the plane, ˆn P is the normal to the plane, and x is any point in space lying on the plane. Abbreviating the curved ray expression 3 as xs = x 0 + ˆn 0 fs + ˆn 0 ˆα gs, 9 where fs is equation 6 and gs is equation 7, the equation for intersection of the curved ray with the plane is x 0 x P ˆn P + ˆn 0 ˆn P fs + ˆn 0 ˆα ˆn P gs = 0 30 This equation is amenable to solution via recursive estimation. Note that, in the special case of no bending = 0, fs = s and gs = 0, and the intersection equation reduces to the one normaliy used for straight ray intersection: s = x P x 0 ˆn P /ˆn 0 ˆn P. Because of this, and because many of the situations of interest involve small amounts of ray bending over long distances, it can be useful to rewrite the curved ray intersection equation to look like the straight ray intersection with an alteration: s = x P x 0 ˆn P fs s + ˆn 0 ˆα ˆn P gs 31 This form is still the full curved ray problem. No approximations have been made to the intersection equaiton, just a rearrangement of terms. This arrangement naturally provides an iterative estimate procedure s n+1 = x P x 0 ˆn P and a natural initialization fs n s n + ˆn 0 ˆα ˆn P gs n 3 = x P x 0 ˆn P 33 This iterative scheme has theoretically unknown convergence properties, but in some preliminary testing in naminal conditions it converges to double precision accuracy in as little at two iterations, and under more stressing conditions in as many at 10 iterations. A special case that this iterative scheme fails is the situation where the plane is parallel to the initial direction ˆn 0, in which case ˆn 0 ˆn P = 0 and the initialization is poor. Under this condition, the intersection equation is x 0 x P ˆn P + ˆn 0 ˆα ˆn P gs = 0 34 ollowing a similar approach as with the other case, an iterative scheme can be set up as with initial value s n+1 = x P x 0 ˆn P ˆn 0 ˆα ˆn P gs n s n + 35 = x P x 0 ˆn P ˆn 0 ˆα ˆn P 36 Preliminary tests show the convergence of this case to be slower, around 100 iterations. 6